An excited He+ion emits two photons in succession,

Question:

An excited $\mathrm{He}^{+}$ion emits two photons in succession, with wavelengths $108.5 \mathrm{~nm}$ and $30.4 \mathrm{~nm}$, in making a transition to ground state. The quantum number $n$, corresponding to its initial excited state is (for photon of wavelength $\lambda$,

energy $\mathrm{E}=\frac{1240 \mathrm{eV}}{\lambda(\text { innm })}$

  1. (1) $n=4$

  2. (2) $n=5$

  3. (3) $n=7$

  4. (4) $n=6$


Correct Option: , 2

Solution:

(2) $E=E_{1}+E_{2}$

$13.6 \frac{z^{2}}{n^{2}}=\frac{1240}{\lambda_{1}}+\frac{1240}{\lambda_{2}}$

or $\frac{13.6(2)^{2}}{n^{2}}=1240\left(\frac{1}{108.5}+\frac{1}{30.4}\right) \times \frac{1}{10^{-9}}$

On solving, $n=5$

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